class MyClass
{
int x, y;
void foo() volatile {
// do stuff with x
// do stuff with y
}
};
Do I need to declare
The following code:
#include <iostream>
class Bar
{
public:
void test();
};
class Foo
{
public:
void test() volatile { x.test(); }
private:
Bar x;
};
int main()
{
Foo foo;
foo.test();
return 0;
}
Raises an error upon compilation with gcc:
main.cpp: In member function 'void Foo::test() volatile':
main.cpp:14:33: error: no matching function for call to 'Bar::test() volatile'
main.cpp:7:8: note: candidate is: void Bar::test() <near match>
And since a volatile
instance can't call a non-volatile
method, we can assume that, yes, x
and y
will be volatile
in the method, even if the instance of MyClass
is not declared volatile
.
Note: you can remove the volatile
qualifier using a const_cast<>
if you ever need to; however be careful because just like const
doing so can lead to undefined behavior under some cases.
Marking a member function volatile
is like marking it const
; it means that the receiver object is treated as though it were declared as a volatile T*
. Consequentially, any reference to x
or y
will be treated as a volatile
read in the member function. Moreover, a volatile
object can only call volatile
member functions.
That said, you may want to mark x
and y
volatile
anyway if you really do want all accesses to them to be treated as volatile
.
You don't have to declare the member variables explicitly..
From Standard docs 9.3.2.3,
Similarly, volatile semantics (7.1.6.1) apply in volatile member functions when accessing the object and its nonstatic data members.
So using the original example:
class MyClass
{
int x, y;
void foo() volatile {
// do stuff with x
// do stuff with y
// with no "non-volatile" optimization of the stuff done with x, y (or anything else)
}
void foo() {
// do stuff with x
// do stuff with y
// the stuff done with x, y (and anything else) may be optimized
}
};